I also need help on this one too.

[tex]\\ \sf\longmapsto tan60=\dfrac{a}{b}[/tex]
[tex]\\ \sf\longmapsto \sqrt{3}=\dfrac{3}{b}[/tex]
[tex]\\ \sf\longmapsto b=\dfrac{3}{\sqrt{3}}[/tex]
[tex]\\ \sf\longmapsto b=\sqrt{3}[/tex]
[tex]\\ \sf\longmapsto b=1.732[/tex]
Now
[tex]\\ \sf\longmapsto a^2+b^2=c^2[/tex]
[tex]\\ \sf\longmapsto (\sqrt{3})^2+3^2=c^2[/tex]
[tex]\\ \sf\longmapsto c^2=9+3[/tex]
[tex]\\ \sf\longmapsto c^2=12[/tex]
[tex]\\ \sf\longmapsto c=3.3[/tex]
Step-by-step explanation:
Solution,
Taking angle A=60° as reference angle, we have
perpendicular(p)=3
hypotenuse(h)=c
base(b)=b
Now,
Tan 60°=p/b
√3=3/b
b=3/√3
b=1.732
Again,
Sin 60°=p/h
√3/2=3/c
√3×c=3×2
√3×c=6
c=6/√3
c=3.4